Question

The function \( f(x) = \begin{cases} (1+2x)^{1/x}, & x \neq 0 \\ e^2, & x=0 \end{cases} \) is
 

• A. Differentiable at x = 0
• B. Continuous at x = 0
• C. Discontinuous at x = 0
• D. Not differentiable at x = 0

Hint:

For limits of the form \( 1^\infty \), i.e., \( \lim (1+f(x))^{g(x)} \) where \( f(x) \to 0 \) and \( g(x) \to \infty \), the result is always \( e^{\lim (f(x) \cdot g(x))} \). This is a quick way to solve such limits, which are common when checking continuity of exponential functions.

Answer 1

The Correct Option is B

To check for continuity at \( x=0 \), we need to evaluate the limit of \( f(x) \) as \( x \to 0 \) and see if it equals \( f(0) \). We are given \( f(0) = e^2 \). Now, let's find the limit: \[ L = \lim_{x \to 0} (1+2x)^{1/x} \] As \( x \to 0 \), the base \( (1+2x) \to 1 \) and the exponent \( 1/x \to \infty \). This is the indeterminate form \( 1^\infty \). We use the standard formula for this form: If \( \lim_{x \to a} g(x) = 1 \) and \( \lim_{x \to a} h(x) = \infty \), then \( \lim_{x \to a} g(x)^{h(x)} = e^{\lim_{x \to a} (g(x)-1)h(x)} \). Here, \( g(x) = 1+2x \) and \( h(x) = 1/x \). \[ L = e^{\lim_{x \to 0} ((1+2x)-1) \cdot \frac{1}{x}} \] \[ L = e^{\lim_{x \to 0} (2x) \cdot \frac{1}{x}} \] \[ L = e^{\lim_{x \to 0} 2} = e^2 \] Since the limit \( \lim_{x \to 0} f(x) = e^2 \) is equal to the value of the function at that point, \( f(0) = e^2 \), the function is continuous at \( x=0 \). While it may or may not be differentiable, it is definitely continuous.